Pure Physics Nov 2016 TYS ANSWERS

Nov 2016_Suggested ANSWERS_as at 1 Jan 2019 Page 1 of 6 – Nov 2016 Catholic High School | O-Level Physics NOT IN SYLLABUS: 5059 Nov 2016 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 C 11 B 21 B 31 D 2 D 12 A 22 C 32 D 3 C 13 B 23 D 33 D 4 C 14 C 24 A 34 C 5 A 15 C 25 B 35 B 6 A 16 B 26 A 36 D 7 C 17 A 27 D 37 A 8 B 18 D 28 B 38 D 9 B 19 A 29 C 39 B 10 C 20 D 30 C 40 C *Q. 1: C (B is incorrect.) *Q. 8: B The distribution of material within the bottle ensured that the centre of mass is originally at point Y. However, the extra sand sinks to a level that is above the original position of the centre of mass and so the centre of mass rises. (C is incorrect.) *Q. 13: B A question that concerns kinetic energy does not always require the answer obtained to be squared. (D is incorrect.) *Q. 32: D Option B: The resistance of the ammeter was not equal to that of the lamps. Option C: There should be less current in the branch with the greater resistance. (B and C are incorrect.) *Q. 36: D The compass needle is needed to retain its magnetised state in order to reveal the direction of the magnetic field. (C is incorrect.) *Q. 37: A The coil reaches equilibrium in a vertical position. (B is incorrect.) *Q. 39: B Although a transformer requires an alternating supply, the a.c. is not converted to d.c. at any stage. (C is incorrect.) *Q. 40: C The current in Y does not decrease in order for the power supplied to stay the same. (D is incorrect.) Nov 2016_Suggested ANSWERS_as at 1 Jan 2019 Page 2 of 6 – Nov 2016 Paper 2 [80 marks] 1 a 10 m s-2, downwards 10 m s-2, downwards 1 1 b 2 ci The displacement-time graph is a curve, which shows that the velocity (gradient of the displacement-time graph) is changing. 1 cii Velocity: constant Acceleration: zero 1 2 a 2 bi a = ௩ ି ௨ ௧ = ଵ.ହ ି ଴ ଴.ହ଴ = 3.00 m s-2 FR = ma = (60)(3.00) = 180 N (3 s.f.) 1 1 bii Area, A = 500 cm2 = 0.0500 m2 W = mg = (60)(10) = 600 N P = ி ஺ = ଺଴଴ ଴.଴ହ଴଴ = 12 000 Pa (3 s.f.) 1 1 biii The force exerted by the man on the ground increases to be larger than his weight (so that there is a resultant force acting on himself upwards). 1 3 a Work is the product of the force applied and the distance moved in the direction of the force. 1 bi Density,  = ௠ ௏ m = V = (1000)(200) = 200 000 kg W = mg = (200 000)(10) = 2 000 000 N Work done = (F)(s) = (2 000 000)(4.0) = 8 000 000 = 8.00  106 J (3 s.f.) 1 1 Note: Velocity is zero at t = 0.4 s weight weight normal reaction force

Nov 2016_Suggested ANSWERS_as at 1 Jan 2019 Page 3 of 6 – Nov 2016 bii P = Work done Time = ଼.଴ × ଵ଴ల 5.0 = 1.6  106 W (3 s.f.) 1 1 c Some of the input power may be lost as thermal and sound energies (due to friction in the pump). 1 4 a Condensation is a change in state of a substance from gaseous to liquid state, without a change in temperature. 1 b When the steam condenses, the seawater gains thermal energy. This causes the kinetic energy of the seawater molecules to increase. Thus the internal energy of seawater increases (= K.E. + P.E. of molecules). 1 1 1 c Q = mc m = ொ ௖∆ఏ = ଶଶ଴ × ଵ଴ల (ଷଽ଴଴)(ସଽ ି ଶ଼) = 2690 kg (3 s.f.) 1 1 5 ai The focal length is the distance between the optical centre of the lens and the focal length. 1 aii [Note: The object must be at larger than 2f.] 3 bi 2 bii The larger the refractive index, the smaller the focal length. With a larger refractive index, the light would bend more towards the normal when it enters glass, and bend more away from the normal when it exits into air. 1 1 Nov 2016_Suggested ANSWERS_as at 1 Jan 2019 Page 4 of 6 – Nov 2016 6 ai 2 aii The direction of the electric field is the direction of the force acting on a positive electrical charge. 1 bi Negative Positive 1 bii The charges on the bottom of the metal strip is closer to the sphere and hence the force is larger. 1 7 a Vout = ோresistor ோresistorା ோLDR  Vtotal = ଺଴଴ ଺଴଴ ା ଼଴଴଴  12 = 0.837 V (3 s.f.) 1 1 bi V across fixed resistor = Vtotal - Vout = 12 - 8 = 4 V (0 d.p.) 1 bii ௏resistor ௏LDR = ோresistor ோLDR  ସ ଼ = ଼଴଴଴ ோLDR RLDR = 8000  ଼ ସ = 16 000  (3 s.f.) 1 c Vout before swappng Vout after swapping Bright 0.837 V 12 - 0.837 = 11.2 V Dark 8.0 V 12 - 8 = 4 V In bright light, the new Vout (across the fixed resistor) is 11.2 V, and the lamp is switched on. As the level of light falls, the new Vout decreases, and the lamp starts to dim until it switches off (when Vout = 4 V). 1 1 8 a Taking moments about the pivot, Clockwise moment = Anticlockwise moments F(25) = 11 + 30 F = 1.64 N (3 s.f.) 1 1 b When a larger current flows in the circuit, the steel core is magnetized and attracts the iron arm, causing it to rotate anticlockwise about the pivot. 1 c The core is made of steel, which is a hard magnetic material and is difficult to demagnetise. Even after the fault in the circuit is rectified, the steel core will still be magnerised and will attract the contacts. 1 9 ai A high-pitched sound is a sound that has a high frequency. 1 aii As the car reverses towards the wall, the beeps increase as distance decreases. The number of beeps is constant at 4 from 140 – 100 cm, then 10 beeps from 80 – 40 cm and finally 20 beeps from 20 – 0 cm. 1 1 aiii Device B is more sensitive (or gives a better indication of the distance from the wall), as the no. of beeps increases generally with every 20 cm. Device A gives the same number of beeps over a larger distance (e.g. 4 beeps over 40 cm, from 140 – 100 cm). 1 1 bi When d = 100 cm  4 beeps per second 1

Nov 2016_Suggested ANSWERS_as at 1 Jan 2019 Page 5 of 6 – Nov 2016 * [Note: from the question, each sound lasts for 0.050 s.] Total time the buzzer produces sound = 4  0. 050 = 0.200 s Hence, fraction of time the buzzer produces sound in each second = 0.200 (of 1 s) = ଵ ହ [Note: The answer has to be expressed as a fraction, as required by the question.] bii 1. v = 2s t  2s = vt = (340)(40  10-3) s = 0.680 m = 68.0 cm (3 s.f.) 1 1 2 From the table for device A: Number of beeps per second = 10 1 c 3 10 ai Fig. 10.1 is not a straight line graph passing through the origin. 1 aii As the temperature of the lamp increases, the resistance of the lamp increases. 1 b From the graph, when V = 6.0 V, I = 35 mA R = ௏ ூ = 6 35 × ଵ଴షయ = 171  (3 s.f.) 1 1 ci E.m.f. is the work done by the source in driving a unit charge around a complete circuit. 1 cii 2 ciii Current in both the lamp and resistor = 35 mA (when the lamp is at 6.0 V) V of resistor = IR = (35  10-3)(200) = 7.00 V E.m.f., E = 6.0 + 7.00 = 13.0 V (1 d.p.) 1 1 11 a 1. Electrocution 1 Note: The straight line must pass through the point (4 V, 20 mA), such that R = ସ ଶ଴ × ଵ଴షయ = 200  Nov 2016_Suggested ANSWERS_as at 1 Jan 2019 Page 6 of 6 – Nov 2016 E 2. Fire caused by damaged insulation, damp conditions or the high voltage or by overloading the socket or circuit. bi I = (no. of lamps)( ௉ ௏) = (26)( ଵଶ ଶଷ଴) + (5)( ସ଴ ଶଷ଴) + (4)( ଵ଼ ଶଷ଴) = 2.54 A (3 s.f.) 2 1 bii 3 A [Note: fuse ratings are in whole no.] 1 ci P = I2R R = ௉ ூమ = ହ.଴ ଺ହమ = 0.00118  (3 s.f.) 1 1 cii 1 P = I2R (P  R) and R =  ௅ ஺ (R  L) Thus power lost is directly proportional to the length of the section. 1 cii 2 P = I2R (P  R) and R =  ௅ ஺ (R  ଵ ஺) Thus power lost is inversely proportional to the cross-sectional area of the section. 1 11 O ai An alternating voltage is a voltage that is changing in magnitude and direction. 1 aii The changing magnetic field around the primary coil causes a change in magnetic flux linkages in the secondary coil, hence inducing a voltage. 1 aiii The iron core concentrates the magnetic flux lin